
If \[f\left( x \right) = {\log _e}\left( {{{\log }_e}x} \right)\], then \[f'\left( e \right)\] is equal to
A. \[{e^{ - 1}}\]
B. e
C. 1
D. 0
Answer
232.8k+ views
Hint: When the limits of a function f(x) exist and function is differentiable at a point such that the slope of the function can be determined at a specific point which can be found by differentiating the function, that is by finding the rate at which the slope of the function is changing and it is denoted by f’(x).
Formula Used: \[\dfrac{{d\left( {{{\log }_e}x} \right)}}{{dx}} = \dfrac{1}{x}\], which is represented by, that the differentiation of \[{\log _e}x\] with respect to x is \[\dfrac{1}{x}\].
Complete step-by-step solution:
We have the given function as \[f\left( x \right) = {\log _e}\left( {{{\log }_e}x} \right)\]. We will first find the derivative of the function with respect to x, which is \[f'\left( x \right)\], and then substitute the value as e in place of x in the derivative of the function.
We will apply the chain rule to find the derivative of the function as,
\[
f\left( x \right) = {\log _e}\left( {{{\log }_e}x} \right) \\
\Rightarrow f'\left( x \right) = \dfrac{{d\left( {{{\log }_e}\left( {{{\log }_e}x} \right)} \right)}}{{dx}} \\
\Rightarrow f'\left( x \right) = \dfrac{1}{{{{\log }_e}x}} \times \dfrac{{d\left( {{{\log }_e}x} \right)}}{{dx}} \\
\Rightarrow f'\left( x \right) = \dfrac{1}{{{{\log }_e}x}} \times \dfrac{1}{x} \\
\]
Further Simplifying, we get,
\[f'\left( x \right) = \dfrac{1}{{x{{\log }_e}x}}\]
Now, we will substitute the value e in place x in \[f'\left( x \right)\] and get the value of \[f'\left( e \right)\] as,
\[
f'\left( x \right) = \dfrac{1}{{x{{\log }_e}x}} \\
\Rightarrow f'\left( e \right) = \dfrac{1}{{e{{\log }_e}e}} \\
\Rightarrow f'\left( e \right) = \dfrac{1}{e} \\
\Rightarrow f'\left( e \right) = {e^{ - 1}} \\
\]
So, option A \[{e^{ - 1}}\] is correct.
Additional Information: Chain Rule is used in differentiation when there is a function inside another function and that means that, while differentiating the function, we have to multiply the derivative of the function which is inside of another function after differentiating another function or the function which is outside.
Note: When we differentiate any function, always check the terms in the function whether they are the variables of the given function or some other constant of another variable and not include them while differentiating. Always remember to apply the chain rule while differentiating a function if there is a function inside another function. When we replace a value with a variable make sure that the other constants remain the same as they were before.
Formula Used: \[\dfrac{{d\left( {{{\log }_e}x} \right)}}{{dx}} = \dfrac{1}{x}\], which is represented by, that the differentiation of \[{\log _e}x\] with respect to x is \[\dfrac{1}{x}\].
Complete step-by-step solution:
We have the given function as \[f\left( x \right) = {\log _e}\left( {{{\log }_e}x} \right)\]. We will first find the derivative of the function with respect to x, which is \[f'\left( x \right)\], and then substitute the value as e in place of x in the derivative of the function.
We will apply the chain rule to find the derivative of the function as,
\[
f\left( x \right) = {\log _e}\left( {{{\log }_e}x} \right) \\
\Rightarrow f'\left( x \right) = \dfrac{{d\left( {{{\log }_e}\left( {{{\log }_e}x} \right)} \right)}}{{dx}} \\
\Rightarrow f'\left( x \right) = \dfrac{1}{{{{\log }_e}x}} \times \dfrac{{d\left( {{{\log }_e}x} \right)}}{{dx}} \\
\Rightarrow f'\left( x \right) = \dfrac{1}{{{{\log }_e}x}} \times \dfrac{1}{x} \\
\]
Further Simplifying, we get,
\[f'\left( x \right) = \dfrac{1}{{x{{\log }_e}x}}\]
Now, we will substitute the value e in place x in \[f'\left( x \right)\] and get the value of \[f'\left( e \right)\] as,
\[
f'\left( x \right) = \dfrac{1}{{x{{\log }_e}x}} \\
\Rightarrow f'\left( e \right) = \dfrac{1}{{e{{\log }_e}e}} \\
\Rightarrow f'\left( e \right) = \dfrac{1}{e} \\
\Rightarrow f'\left( e \right) = {e^{ - 1}} \\
\]
So, option A \[{e^{ - 1}}\] is correct.
Additional Information: Chain Rule is used in differentiation when there is a function inside another function and that means that, while differentiating the function, we have to multiply the derivative of the function which is inside of another function after differentiating another function or the function which is outside.
Note: When we differentiate any function, always check the terms in the function whether they are the variables of the given function or some other constant of another variable and not include them while differentiating. Always remember to apply the chain rule while differentiating a function if there is a function inside another function. When we replace a value with a variable make sure that the other constants remain the same as they were before.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

